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Pani-rosa [81]
3 years ago
5

Can someone help me on this

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
8 0
155,500



I did this to the best of my ability. I have a hard time comprehending things sometimes so I’m so so so sorry if it’s wrong
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Jack researched J.J. Thomson’s experiments. Read Jack’s summary below and find the error.
WINSTONCH [101]

Answer:

D. Error is in sentence 1. The beam was a cathode ray.

Explanation:

Legit just looked up his experiments and it said right off the bat he studied cathode rays not gamma.

8 0
3 years ago
__________ is considered to be the greatest solvent in the world because it can dissolve so many substances.
allochka39001 [22]
The correct answer is c. water. 
6 0
3 years ago
1) How many Joules of energy are released when 75g of water is heated
Pie

Answer: 23,199 J

Explanation:

8 0
2 years ago
Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
3 years ago
Write the balanced neutralization reaction that occurs between h2so4 and koh in aqueous solution. Phases are optional.
Nutka1998 [239]

Answer:

The answer to your question is below

Explanation:

H₂SO₄ = sulphuric acid

KOH = potassium hydroxide

This is a neutralization reaction

                       H₂SO₄   +  2 KOH   ⇒     K₂SO₄   +  2H₂O

                                 2 -----------   K  ----------- 2

                                 1 -----------   S  ----------  1

                                 4 ----------   H  ---------   4

                                6 ------------   O ----------  6

4 0
3 years ago
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